Differential EquationsmediumFree

Differential Equations: Solution Differential Equation

JEE Maths question with a full step-by-step solution.

Question

The solution of the differential equation

(x^2\sin^3 y - y^2\cos x)\,dx + (x^3\cos y\sin^2 y - 2y\sin x)\,dy = 0

is:

x^3\sin^3 y = 3y^2\sin x + C

correct

x^3\sin^3 y + 3y^2\sin x = C

x^2\sin^3 y + y^3\sin x = C

2x^2\sin y + y^2\sin x = C

Solution

Group the terms so that each group is an exact differential:

\big(x^2\sin^3 y\,dx + x^3\sin^2 y\cos y\,dy\big) - \big(y^2\cos x\,dx + 2y\sin x\,dy\big) = 0.

Now note

d\left(\frac{x^3}{3}\sin^3 y\right) = x^2\sin^3 y\,dx + x^3\sin^2 y\cos y\,dy,

d\big(y^2\sin x\big) = y^2\cos x\,dx + 2y\sin x\,dy.

So the equation is

d\left(\dfrac{x^3}{3}\sin^3 y\right) - d\big(y^2\sin x\big) = 0

, giving

\frac{x^3}{3}\sin^3 y - y^2\sin x = c \Rightarrow x^3\sin^3 y = 3y^2\sin x + C.

Correct answer: (1)

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